Any Ramanujan fans here?

source: r/mathmemes


no, you’re not infinity, you’re -1/12

Can anyone give me the proof of
1+2+3+... = \frac{-1}{12}

1 Like

@Tahmid_S can you help?

Let S = 1+2+3+4+5+6+7

Consider S_1= 1-1+1-1+1-1+1-1…..

Now, this sum should be 0 or 1 based on number of natural numbers taken. If infinite numbers are even, S_1=0, if odd S_1=1. But, Riemann zeta function gives it a value of \frac 1 2. Mathematical community too agrees that the sum is \frac 1 2.

How? At first instance, it feels like the whole community of mathematicians are playing a prank. It is like celebrating April fools day! But yes, serious mathematical work went into the proof. If you are interested to know, please go through Ramanujan’s summation principles and zeta function. Let me try a simple proof avoiding all the complexity.





So, S_1=\frac 1 2

Several objections can be put like why not the alternative solution of 0 or 1? But as stated before, we need much powerful tools in mathematics like zeta functions to come to unique solution of \frac 1 2. For now, we could agree S_1=\frac 1 2.

Let S_2=1-2+3-4+5-6+7…..

So, S_2=1-2+3-4+5-6+7-8+9…..

S_2= 1-2+3-4+5-6+7-8……. I have shifted RHS by a unit position

+ 2S_2=1-1+1-1+1-1+1…..

Hence, 2S_2=S_1

Therefore, S_2=\frac 1 4

Let’s come back to our sum of infinite numbers.



So, S-S_2=4+8+12+16+20…..

Hence, S-S_2=4(1+2+3+4+5+6+7+8….)


So, -S_2=3S

And, S=\frac{-S_2}{3}= -\frac{1}{12}

Source: Is sum of Natural Numbers upto Infinity -1/12?? | AvantGarde


A proof that 1+2+3+...=\frac {-1}{8}

$\frac {-1}{8}$

Let S = 1+2+3+...
S = 1+2+3+4+5+6+7+...
=> S = 1 + 3×3 + 6×3 + ...
S=> S = 1 + 9(1+2+…) => S = 1 + 9S$
=> -1 = 8S
=> S = \frac {-1}{8}

[Note :- I saw a similar proof like the above about \frac {-1}{8}]

But if we watch carefully , we will find this infinite sum can also be infinity if we put it in
S =1+9S


Note :- this sum is also infinite if we put it instead the value of S in the equation proving the value of \frac {-1}{12}

I am a great fan of Ramanujan :grin: